# ML Aggarwal Algebra Exe-9.5 Class 6 ICSE Maths Solutions

ML Aggarwal Algebra Exe-9.5 Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-9.5 Questions for Algebra as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-6.

## ML Aggarwal Algebra Exe-9.5 Class 6 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 6th Chapter-9 Algebra Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-9.5 Questions Edition 2023-2024

### Algebra Exe-9.5

ML Aggarwal Class 6 ICSE Maths Solutions

Page-198

#### Question 1. State which of the following are equations with a variable. In case of an equation with a variable, identify the variable.

(i) 17 + x = 5
(ii) 2b – 3 = 7
(iii) (y – 7) > 5
(iv) 9/3 = 3
(v) 7 × 3 – 19 = 2
(vi) 5 × 4 – 8 = 31
(vii) 2p < 15
(viii) 7 = 11 × 5 – 12 × 4
(ix) 3/2 q = 5

(i) 17 + x= 5 Is an equation → L.H.S. = R.H.S. → Related variable x.
(ii) 2b – 3 = 7 Is an equation → L.H.S. = R.H.S. → Related variable b.
(iii) (y- 7) >5
Is not an equation → L.H.S. ≠ R.H.S.
It has no sign of equality (=).
(iv) 9/3 = 3
Is an equation = L.H.S. = R.H.S.
It has no variable.
(v) 7 × 3 – 19 = 2
Is an numerical equation = L.H.S. = R.H.S. It has no variable.
(vi) 5 × 4 – 8 = 31
Is an equation = L.H.S. = R.H.S. → Related variable t.
(vii) 2p < 15
Is not an equation = L.H.S. ≠ R.H.S.
It has no sign of equality.
(viii) 7 = 11 × 5 – 12 × 4
Is an numerical equation = L.H.S. = R.H.S. It has no variable.
(ix) 3/2 q = 5
Is an equation → L.H.S. = R.H.S. → Related variable q.

#### Question 2.  Solve each of the following equations :

(i) x + 6 = 8
(ii) 2 – x = 5
(iii) 4x = -6
(iv) x/2 = 5
(v) 2y – 3 = 2
(vi) 4 – 5y = 2

(i) x + 6 = 8
= x = 8 – 6 ⇒ x = 2
(ii) 2 – x = 5
= -x = 5 – 2 ⇒ -x = 3 ⇒ x = – 3
(iii) 4x = -6
$=x=\frac{-6}{4}=\frac{-3}{2}$
(iv) x/2 = 5
= x = 5 × 2 ⇒ x = 10
(v) 2y – 3 = 2
= 2y = 2 + 3 ⇒ 2y = 5 ⇒ y = 5/2
(vi) 4 – 5y = 2
= 4 – 2 = 5y ⇒ 5y = 2
⇒ y = 2/5

#### Question 3.Solve the following linear equations:

(i) 5(x + 1) = 25
(ii) 2(3x – 1) = 10
(iii) (3x – 1)/4 = 11

(i) 5(x + 1) = 25
⇒ 5(x + 1)/5 = 25/5  (dividing both sides by 5)
⇒ x + 1 = 5
⇒ x + 1 – 1 = 5 – 1 (Subtracting 1 from both sides)
⇒ x = 4

(ii) 2(3x – 1) = 10
⇒ 2(3x – 1)/2 = 10/2  (dividing both sides by 2)
⇒ 3x – 1 = 5
⇒ 3x – 1 + 1 = 5 + 1 (adding 1 to both sides)
⇒ 3x = 6
⇒ 3x/3 = 6/3  (dividing both sides by 3)
⇒ x = 2

(iii) Given (3x – 1)/4 = 11
⇒ 4 × (3x – 1)/4 = 4 × 11  (multiplying both sides by 4)
⇒ 3x – 1 = 44
⇒ 3x – 1 + 1 = 44 + 1 (adding 1 to both sides)
⇒ 3x = 45
⇒ 3x/3 = 45/3  (dividing both sides by 3)
⇒ x = 15

#### Question 4. Solve the following linear equations:

(i) 5x – 6 = 12 – x

(ii) n/3 + 1 = 4 – n

(iii) 5p + 7 = 19 – 2p

(iv) 2x + 5/2 = 2/3 -x

(v) x/2 -5 = x/3 -4

(vi) 18 – (3y/4) = 11 + y

(i) 5x – 6 = 12 – x
⇒ 5x + x = 12 + 6

6x = 18 ⇒ x  = 18/6 = 3

Verification
5x − 6 = 12 − x

⇒ 5(3) − 6 = 12 − 3
⇒ 15 – 6 = 9
⇒ 9 = 9

(ii) n/3 + 1 = 4 – n

(iii) 5p + 7 = 19 – 2p

⇒ 5p + 2p = 19 – 7

⇒ 7p = 12

⇒ p 12/7

Verification

(iv) 2x + 5/2 = 2/3 -x

(v) x/2 – 5 = x/3 -4

(vi) 18 – (3y/4) = 11 + y

(i) 3(x + 7) = 18
(ii) 2(x- 1) = x + 2
(iii) 3x – 1/2 = 2( x – 1/2) + 5
(iv) 4(2x – 1) -2(x – 5) = 5(x + 1) + 3

(i) 3(x + 7) = 18
⇒ 3x + 21 = 18
⇒ 3x = 18 − 21
⇒ 3x = −3
⇒ x = -3/3
x = −1
Verification
⇒ 3(x + 7) = 18
⇒ 3(−1 + 7) = 18
⇒ 3(6) = 18
⇒ 18 = 18

(ii) 2(x − 1) = x + 2
⇒ 2x − 2 = x + 2
⇒ 2x − x = 2 + 2
⇒ x = 4
Verification
⇒ 2(x − 1) = x + 2
⇒ 2(4 − 1) = 4 + 2
⇒ 2(3) = 6
⇒ 6 = 6

(iv) 4(2x – 1) -2(x – 5) = 5(x + 1) + 3

⇒ 8x − 4 − 2x + 10 = 5x + 5 + 3
⇒ 8x − 2x− 4 + 10 = 5x + 5 + 3
⇒ 6x + 6 = 5x + 8
⇒ 6x – 5x = 8 – 6
⇒ x = 2
Verification :

4(2x − 1) − 2(x − 5) = 5(x + 1) + 3
⇒ 4 (2 × 2-1) – 2(2-5)=5(2+1)+3
⇒ 4 ( 4 – 1) – 2 (-3) = 5 ( 3) + 3
⇒ 4 ( 3) – 2 (-3) = 15 + 3
⇒ 12 + 6 = 18
⇒ 18 = 18

—  : End of ML Aggarwal Algebra Exe-9.5 Class 6 ICSE Maths Solutions :–

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